You have found the following ages (in years) of all 4 meerkats at your local zoo: $ 1,\enspace 16,\enspace 2,\enspace 7$ What is the average age of the meerkats at your zoo? What is the variance? You may round your answers to the nearest tenth.
Solution: Because we have data for all 4 meerkats at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $4$ ages and divide by $4$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{1 + 16 + 2 + 7}{{4}} = {6.5\text{ years old}} $ Find the squared deviations from the mean for each meerkat. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $1$ year $-5.5$ years $30.25$ years $^2$ $16$ years $9.5$ years $90.25$ years $^2$ $2$ years $-4.5$ years $20.25$ years $^2$ $7$ years $0.5$ years $0.25$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{30.25} + {90.25} + {20.25} + {0.25}} {{4}} $ $ {\sigma^2} = \dfrac{{141}}{{4}} = {35.25\text{ years}^2} $ The average meerkat at the zoo is 6.5 years old. The population variance is 35.25 years $^2$.